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    • 4. 发明专利
    • Ti-Cu-Zr-Pd METAL GLASS ALLOY
    • Ti-Cu-Zr-Pd金属玻璃合金
    • JP2009052131A
    • 2009-03-12
    • JP2008043996
    • 2008-02-26
    • Tohoku Univ国立大学法人東北大学
    • INOUE AKIHISAWANG XINMINAKE KATSUTOSHIHATA FUGAWADA TAKESHI
    • C22C45/10C22C45/00
    • PROBLEM TO BE SOLVED: To provide a Ti-Cu-Zr-Pd metal glass alloy having excellent bulk formability. SOLUTION: The Ti-Cu-Zr-Pd metal glass alloy has a composition expressed by formula: Ti 100-a-b-c Cu a Zr b Pd c [wherein, a, b and c are atomic%, and, a denotes 30 to 50%, b denotes 0.5 to 20% and c denotes 0.5 to 20%]. Alternatively, the Ti-Cu-Zr-Pd metal glass alloy has a composition expressed by formula: ä(Ti 1-a-b Zr a Nb b ) 0.5 (Cu 1-c-d Pd c Ag d ) 0.5 } 100-x Si x [wherein, a, b, c and d are atomic ratio, and, a denotes 0.005 to 0.2, b denotes 0.005 to 0.1, c denotes 0.005 to 0.2 and d denotes 0.005 to 0.1, and x is an atomic%, and denotes 0.5 to 10%]. COPYRIGHT: (C)2009,JPO&INPIT
    • 要解决的问题:提供具有优异的堆积成形性的Ti-Cu-Zr-Pd金属玻璃合金。 < P>解决方案:Ti-Cu-Zr-Pd金属玻璃合金具有由下式表示的组成:Ti 100-abc 其中a,b和c为原子%,a为30〜50%,b为0.5〜20%,c为0.5〜20%。 或者,Ti-Cu-Zr-Pd金属玻璃合金具有由以下公式表示的组成:ä(Ti 1-ab b B < SB> 0.5< SB>(Cu< SB> 1-cd< SB>< SB> 其中a,b,c和d是原子比,a表示0.005〜0.2,b表示0.005〜0.1,c表示0.005 为0.2,d为0.005〜0.1,x为原子%,为0.5〜10%]。 版权所有(C)2009,JPO&INPIT